Given a positive integer $n$, a group $G$ of order $n$, and a divisor $d$ of $n$, in what cases can we be assured of the existence of a subgroup $H$ of $G$ of order $d$?

What's the situation in the case of the symmetric group on $m$ letters or for the alternating group?

What's the most general statement in each case?

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    $\begingroup$ See also this question. In terms of $d$, we cannot do any better than Sylow's theorem. If $d$ is not a power of a prime, then there exists a group of order divisible by $d$ but with no subgroup of order $d$. In terms of $G$, the groups that satisfy the converse of Lagrange's theorem are called CLT groups. You can show that supersolvable groups are CLT and that CLT groups are solvable. $\endgroup$ – Mikko Korhonen Feb 11 '13 at 15:51

See these two links to have additional information:

The inverse of Lagrange's Theorem is true for finite supersolvable group.


A kind of converse of Lagrange's Theorem. In the second one, there is a great classification for the problem done by missed @Arturo Magidin.

  • $\begingroup$ Helpful links! +1 $\endgroup$ – amWhy Feb 12 '13 at 0:02

Sylow's theorems deal with the case when $d$ is a prime power. One can show that if $d$ is not a prime power, then there's a multiple $n$ of $d$ and a group $G$ of order $n$ with no subgroups of order $d$.

Here's a reference (please let me know if it is not accessible): Donald McCarthy, Sylow's theorem is a sharp partial converse to Lagrange's theorem. Math. Z. 113 (1970) 383-384.


Two general theorem are:

Cauchy's Theorem: If $G$ is a finite group and $p$ is a prime number dividing $|G|$ then $G$ has an element of order $p$ (and thus a subgroup of order $p$).

Every $p$ group $G$ has a subgroup of any order dividing $|G|$.


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